n equally spaced points are taken on the circumference of a circle of radius 1. How do you find the perimeter of the resulting regular polygon obtained by joining the n points in order?
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$\begingroup$ Connect the vertices to the center. Now you have $n$ equal isosceles triangles. How do you find the lengths of their bases? $\endgroup$– player3236Commented Sep 21, 2020 at 17:39
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3 Answers
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Hint:
Partition the regular polygon in $n$ isosceles triangles with a common vertex at the centre of the circle. The altitude of such a triangle is equal to $\cos\frac{\pi}n$. Can you find its base?
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Answer: $n \sqrt{2r^2 (1-\cos(2\pi/n))}$. This formula is equivalent to:
$$ 2nr \sin(\pi/n) $$
Explanation: Using the cosine rule: since we are given 2 sides and the opposite angle of the missing side, we obtain the last side with the following formula: $$ c^2 = a^2 + b^2 - 2ab \cos(\theta)$$ $$ \implies \text{missing}^2 = 2r^2 - 2r^2 \cos(2\pi/n) $$
The equivalence to the second formula comes from the fact that $$ \sqrt{2-2\cos(2x)} = 2\sin(x) $$
width = 100pxHonestly, I don't know why people on this website are unable to provide quick answers to easy problems. I had this same question, and even though it only took me like 5 minutes to derive the formula, this is the kind of stuff you would actually prefer a formula answer instead of a vague hint.
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Bisect one triangle of the polygon by a radial line. Angle at circle center is $ \dfrac {2\pi}{2n}. $
We see a right angled triangle of sides:
$$ ( 1, \cos \frac {\pi}{n}, \sin \frac {\pi}{n})$$
Directly the polygon perimeter is $$ 2 n \sin \frac {\pi}{n} $$
Give it a check:
$$\lim_{~~n\to \infty} 2 n \sin \dfrac {\pi}{n} = 2\pi$$
answered Dec 2, 2023 at 20:02